How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola #x^2/81-y^2/49=1#?

1 Answer
Nov 20, 2016

Answer:

The vertices are #(9,0)# and #(-9,0)#
The foci are F#(sqrt130,0)# and F'#(-sqrt130,0)#
The asymptotes are #y=7/9x# and #y=-7/9x#

Explanation:

Let's compare this equation to the gerenal equation of a left-right hyperbola

#(x-h)^2/a^2-(y-k)^2/b^2=1#

#x^2/81-y^2/49=1#

The center is #(h,k)=(0,0)#

The vertices are #(h+-a,k)=(+-9,0)#

The slope of the asymptotes are #+-b/a=+-7/9#

The equations of the asymptotes are #y=b/a(x-h)#

The asymptotes are #y=7/9x# and #y=-7/9x#

To determine the foci, we need #c=+-sqrt(a^2+b^2)#

#c=+-sqrt(81+49)=sqrt130#

The foci are F#=(h+c,k)=(sqrt130,0)#

and F'#=(h-c,k)=(-sqrt130,0)#

graph{(x^2/81-y^2/49-1)(y-7/9x)(y+7/9x)=0 [-20.27, 20.29, -10.14, 10.13]}